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\section{AVF referral policy for the on-HD period}
Consider a patient who is under hemo-dialysis (HD), does not have an AVF access, and has at least one AVF chance left, and thus AVF creation is under consideration at the moment. We want to investigate the optimal AVF referral policy. We consider two different objective functions, patient's expected total lifetime. and patient's quality adjusted life expectation (QALE), and derive optimal policies analytically for each.

\section{Expected total lifetime}

Hemo-dialysis using AVF has a better survival than CVC. Nevertheless, empirical data shows that the AVF's superiority of survival over CVC diminishes as patients age. As a consequence, one may think that we should utilize AVF chances at patient's younger ages. In Theorem \ref{thm:main}, we formally state this intuitive idea and then prove it under some assumptions that will later follow.

\begin{thm} \label{thm:main}
Under Assumptions \ref{ass:dec}-\ref{ass:converging}, delaying AVF referral stochastically decreases a patient's life-time.
\end{thm}
\begin{cor}
The optimal policy to maximize patient's expected total lifetime is to refer patient for AVF creation as soon as a patient starts HD (if not referred earlier), or once an existing AVF fails.
\end{cor}
\noindent Before stating our assumptions, we introduce some notation. In what follows, by residual lifetime at time $t$, we mean remaining lifetime from time $t$ onward conditional on survival until time $t$.

\begin{itemize}
\item $C(t)$: R.V. denoting patient's residual lifetime on CVC at time $t$
\item $A(t)$: R.V. denoting patient's residual lifetime on AVF at time $t$
\item $g(t)$: pdf of $C(0)$ at time $t$
\item $\overline{G}(t)$: survival function of $C(0)$ at time $t$
\item $h(t)$: pdf of $A(0)$ at time $t$
\item $\overline{H}(t)$: survival function of $A(0)$ at time $t$
\end{itemize}
Note that the patient's age and the variable $t$ is calculated with respect to some certain age reference, e.g. age of 40 years. Later, we prove that the time origin can be set arbitrarily for drawing conclusions about residual lifetime.\\


Our first assumption is about using an AVF until its failure time. This assumption dictates the decision time-points.
\begin{assu} [Decision points] \label{ass:dec}
A patient cannot start an AVF creation while an AVF is under preparation or when it is functional; in other words, the only time AVF creation is under consideration is when a patient has just started HD with CVC, or when a working AVF has just failed. Although it might be optimal to create a new AVF when the old one is approaching the end of its lifetime, this is not clinically realistic and thus not considered here.
\end{assu}

Our second assumption is about the stochastic process that governs patient's lifetime under HD:

\begin{assu} [Survival distribution] \label{ass:surv}
A patient's survival distribution depends only on age (we do not consider the effect of history of disease on survival). For instance, the 1-year survival probability of a 60 year old patient who started HD at age 55 is considered to be the same as a 60 year old patient who has just started HD. Numerically we have the following equalities:
\begin{align}
&\Pr(A(t) > s) = \Pr(A(0) > t+ s | A(0)>t],\label{eq:A} \\
&\Pr(C(t) > s) = \Pr(C(0) > t+s | C(0)>t],\label{eq:B}
\end{align}
and that $A(t) \perp C(t')$, $A(t) \perp A(t')$, and $C(t) \perp C(t')$ for all meaningful values of $t$, and $t'$, where $\perp$ denotes the independence of random variables.
\end{assu}

Before stating the next assumption, we give the definition of hazard rate order for random variables:
\begin{defn} [Hazard rate order]
We say $X \le_{hr} Y$, if and only if
\begin{align*}
\frac{g(t)}{\overline{G}(t)} \le \frac{f(t)}{\overline{F}(t)}: \forall t.
\end{align*}
where $X$, and $Y$ are random variables with survival functions $\Fbar$, and $\Gbar$, and probability density functions $f$, and $g$, respectively.
\end{defn}


The following two assumptions describe the relative performance of HD with CVC versus AVF, and its dynamics over time. Empirical data confirms these assumptions as well.
\begin{assu} [Relative performance]  \label{ass:relative}
A patient's residual lifetime under HD with CVC is stochastically  smaller than lifetime under AVF (in the usual sense), at all ages. In mathematical words,
\begin{align*} 
 C(t) \le_{st} A(t) , \forall t
\end{align*}
Note that according to Theorem (1.B.7) of \cite{shaked2007stochastic}, this assumption is equivalent to $ C(0) \le_{hr} A(0)$, which implies:
\begin{align} \label{eq:rel}
\frac{h(t)}{\overline{H}(t)} \le \frac{g(t)}{\overline{G}(t)}: \forall t.
\end{align}
 \end{assu}

\begin{assu} [Converging performance] \label{ass:converging}
The difference in performance of HD on AVF versus CVC diminishes as patient ages. More specifically, the difference of hazard rates of $C(0)$ and $A(0)$ is decreasing in time, i.e. 
\begin{align} \label{eq:conv}
\frac{g(t)}{\overline{G}(t)} - \frac{h(t)}{\overline{H}(t)} \downarrow t.
\end{align}
\end{assu} 

\begin{note} \label{note:origin}
The hazard rate functions of $A(t)$ and $C(t)$ at time $s$ (with respect to $t$) are $\hr{h}{t+s}$, and $\hr{g}{t+s}$, respectively. As a result of equations \ref{eq:rel}, and \ref{eq:conv}, we have that Assumptions \ref{ass:relative}-\ref{ass:converging} also hold for $A(t)$ and $C(t)$. This result is important in establishing results about the residual lifetime, as the time origin can be set arbitrarily.
\end{note}

\subsection{Stochastic order of referral policies}

In this section, we prove Theorem \ref{thm:main} indirectly as follows. We prove that the residual lifetime of the patient decreases stochastically in the time we aim to attempt to use an AVF, instead of the time we refer patients for AVF preparation. Because of the surgery waiting and also the maturation time, we cannot choose to use AVF at an arbitrary time; nevertheless, using the fact that any delay in referral directly delays the AVF attempt time, we can extend the result of AVF attempt time to the AVF creation referral time. The following variables will be used in what follows:

\begin{itemize}
\item $u$: time to attempt using an AVF (decision variable) 
\item $L(t,n)$: patient's residual lifetime at time $t$, under optimal AVF referral policy, given $n$ remaining AVF chances
\item $L(t,n,u)$: patient's residual lifetime at time $t$, when we attempt to use AVF at $t+u$, given $n$ remaining AVF chances 
\item $p$: AVF creation success probability 
\item $B_i$: R.V. indicating whether the $i$th last AVF creation is successful
\item $K_i$: R.V. denoting the $i$th last AVF's lifetime (if AVF creation is successful, i.e. $B_i=1$)
\end{itemize}


The next lemma proves that with the goal of maximizing the probability of survival until a certain point, the optimal decision is to use an AVF at the beginning of the time-interval.
\begin{lem}\label{lem:sequence}
Consider Figure \ref{fig:fig3}, and assume that the lifetime of AVF is deterministic and equal to some arbitrary value $k$. Define $S(u,a)$ as the probability of survival until some arbitrary time $a$ when we switch to AVF at time $u$ (for $k\le a$, and $0 \le u \le a-k$). Then, under Assumption \ref{ass:relative}-\ref{ass:converging} the function $S(u,a)$ is decreasing in $u$.
\end{lem}
\begin{figure}[htbp]
\centering
\includegraphics[scale=0.6]{./files/fig3}
\caption{the function $S(u,a)$}
\label{fig:fig3}
\end{figure}

\begin{proof}
~\\
We can calculate $S(u,a)$ as follows:
\begin{align*}
S(u,a)&=\Pr [C(0)>u].\Pr[A(u)>k \big |C(0)>u].\Pr[C(u+k)>a-(u+k) \big |A(u)>k , C(0)>u]\\
&= \overline{G}(u).\frac{\overline{H}(u+k)}{\overline{H}(u)}\frac{\overline{G}(a)}{\overline{G}(u+k)}=
\frac{\dfrac{\overline{G}(u)}{\overline{H}(u)}}{\dfrac{\overline{G}(u+k)}{\overline{H}(u+k)}}.\overline{G}(a).
\end{align*}
where in the second equality, we have used equations \ref{eq:A}-\ref{eq:B}, and the independence of $C(0)$, $A(u)$ and $C(u+k)$ by Assumption \ref{ass:surv}.

Define $y(u)=\dfrac{\overline{G}(u)}{\overline{H}(u)}$, and $z(u)=\dfrac{y(u)}{y(u+k)}$. We need to prove that $z(u)$ is decreasing in $u$. We prove that by showing $z'(u) \le 0$ as follows:
\begin{align*}
z'(u)= \frac{y'(u)y(k+u)-y'(k+u)y(u)}{y^2(k+u)}.
\end{align*}
In order to prove $z'(u) \le 0$, we need to prove that $\dfrac{y'(u)}{y(u)} \le \dfrac{y'(u+k)}{y(u+k)}$. To that end, it suffices to show that $\dfrac{y'(u)}{y(u)}$ is increasing in $u$ as follows:
\begin{align*}
\dfrac{y'(u)}{y(u)}=\dfrac{\dfrac{h(u)\overline{G}(u)-g(u)\overline{H}(u)}{\overline{H}^2(u)}}{\dfrac{\overline{G(u)}}{\overline{H}(u)}}=\frac{h(u)}{\overline{H}(u)} - \frac{g(u)}{\overline{G}(u)},
\end{align*}
which is increasing in $u$ by Assumption \ref{ass:converging}. Note that the result of this lemma extends to the case when the time origin is an arbitrary time $t=t_0$ based on note \ref{note:origin}.
\end{proof}
 The following lemma is about the preservation of stochastic order of random variables and is used in proving the stochastic order of referral policies.
\begin{lem}\label{lem:pres}
Let $X$, $Y$, $Z$ be random variables such that $[X | Z=z] \le_{st} [Y|Z=z]$ for all values of $z$. Then, we have that $X \le_{st} Y$.
\end{lem}

\begin{proof}
We need to prove that $\Pr(X \ge a) \le \Pr (Y \ge a): \forall a$. Note that $\Pr(X \ge a) =\Ex[\Pr(X \ge a|Z=z)]$, and $\Pr(Y \ge a) =\Ex[\Pr(Y \ge a|Z=z)]$, where the expectations are with respect to the random variable $Z$. By assumption, we have $\Pr(X \ge a|Z=z) \le \Pr(Y \ge a|Z=z)$  for all values of $z$, and thus the result follows.
\end{proof}
The next proposition indicates that a patient's lifetime decreases stochastically as the AVF attempt time increases.

\begin{prop}\label{prop:aux}
Under Assumptions \ref{ass:relative}-\ref{ass:converging}, $L(t,n,u_2) \le_{st} L(t,n,u_1)$, whenever $u_1 \le u_2$.
\end{prop}
\begin{proof}
~\\
We prove the theorem by induction on $n$. Note that we can instead prove $L(0,n,u_2) \le_{st} L(0,n,u_1)$ based on note \ref{note:origin}. To that end, we show 
$\big[L(0,n,u_2) \big |B_n=b, K_i=k\big] \le_{st} \big[L(0,n,u_1)\big | B_n=b, K=k\big]: \forall k, b.$
Then using Lemma \ref{lem:pres}, the result follows. Define $S_i(x)=\Pr\big(\big [L(0,n,u_i)\big | B_n=b, K=k\big] > x\big)$. We need to prove $\forall x \ge 0: S_1(x) \ge S_2(x)$.\\

\begin{figure}[htbp]
\centering
\includegraphics[scale=0.6]{./files/fig1}
\caption{the random variable $L(0,1,u)$ and the two possible cases}
\label{fig:fig1}
\end{figure}

\noindent $\rightarrow$ Base case: $n=1$:\\
Consider Figure \ref{fig:fig1}. If AVF creation is unsuccessful ($B_1=0$) (the upper branch in Figure \ref{fig:fig1}), we have that $S_1(x)=S_2(x)=\Pr (C(0) >x)$. Now, assume that the AVF creation is successful. We prove the result by investigating all possible cases for $x$ depicted in Figure \ref{fig:fig4}. In this figure, we have assumed that $u_2 \le u_1+k$. For the case $u_1 +k \le u_2$, the line of proof is the same.

\begin{figure}[htbp]
\centering
\includegraphics[scale=0.6]{./files/fig4}
\caption{possible cases for the induction base case}
\label{fig:fig4}
\end{figure}
\begin{itemize}
\item Case 1 ($x \le u_1$): This case is trivial as $S_1(x)=S_2(x)=\Pr(C(0) > x)$.

\item Case 2 ($u_1 \le x \le u_2$): We have $S_1(x)=\Pr(C(0) > u_1).\Pr(A(u_1) > x-u_1)$, and $S_2(x)=\Pr(C(0) > u_1).\Pr(C(u_1) > x-u_1)$. By Assumption \ref{ass:relative}, $C(u_1) \le_{st
} A(u_1) $ and thus the result follows.

\item Case 3 ($u_2 \le x \le u_1+k$): We have $S_1(x)=S_1(u_2) .\Pr(A(u_2) > x-u_2)$, and $S_2(x)=S_2(u_2). \Pr(A(u_2) > x-u_2)$. As  a result of case 2, we have  $S_2(u_2) \le S_1(u_2)$, and thus $S_2(x) \le S_1(x)$.
\begin{figure}[htbp]
\centering
\includegraphics[scale=0.60]{./files/fig6}
\caption{case 4 and the hypothetical random variable $L_0$}
\label{fig:fig6}
\end{figure}
\item Case 4 ($u_1+k \le x \le u_2+k$): Consider Figure \ref{fig:fig6}. Define $L_0$ a R.V. denoting patient's residual lifetime under a hypothetical scenario similar to $L(0,1,u_1)$ but with the difference that AVF's lifetime is instead $k_0:=u_2+k-x$. Using Lemma \ref{lem:sequence}, we have $S_2(x) \le S_0$, where $S_0$ is the survival probability of random variable $L_0$ until time $x$. It suffices to show $S_0 \le S_1(x)$.
We have
\begin{align*}
S_0=&S_1(k_0).\Pr(C(k_0)>k-k_0).\Pr(C(k)>x-u_1-k),\\
S_1(x)=&S_1(k_0).\Pr(A(k_0)>k-k_0).\Pr(C(k)>x-u_1-k).
\end{align*}

But since $C(k_0) \le_{st} A(k_0)$ by Assumption \ref{ass:relative}, the inequality $S_0 \le S_1(x)$ holds.

\item Case 5 ($u_2+k \le x $):  We have $S_1(x)=S_1(u_2+k).\Pr(C(u_2+k)>x-(u_2+k))$, and $S_2(x)=S_2(u_2+k).\Pr(C(u_2+k)>x-(u_2+k))$. As a result of case 4, we have $S_2(u_2+k) \le S_1(u_2+k)$ and thus the result follows.

\end{itemize}
\noindent $\rightarrow$ Induction step: Assume $L(t,n-1,u_2) \le_{st} L(t,n-1,u_1)$, whenever $u_1 \le u_2$; we prove $L(t,n,u_2) \le_{st} L(t,n,u_1)$.\\
Consider Figure \ref{fig:fig9} and note the following two cases:
\begin{figure}[htbp]
\centering
\includegraphics[scale=0.60]{./files/fig9}
\caption{induction step and the two possible cases}
\label{fig:fig9}
\end{figure}
\begin{itemize}
\item AVF creation fails: Consider Figure \ref{fig:fig10}.


If $x \le u_1$, $S_1(x)=S_2(x)=\Pr(C(0)>x)$. Otherwise, $S_1(x)=\Pr(C(0)>u_1).\Pr(L(u_1,n-1,0) >x-u_1)$, and $S_2(x)=\Pr(C(0)>u_1).\Pr(L(u_1,n-1,u_2-u_1) >x-u_1)$. By induction assumption, we have $L(u_1,n-1,u_2-u_1)  \le_{st} L(u_1,n-1,0)$ and thus the result follows.
\begin{figure}[htbp]
\centering
\includegraphics[scale=0.60]{./files/fig10}
\caption{induction step - AVF creation failure}
\label{fig:fig10}
\end{figure}
\item AVF creation successful: Consider Figure \ref{fig:fig11}. We show the result by showing $L(0,n,u_2) \le_{st} L_0$ and $L_0 \le_{st} L(0,n,u_1)$, where $L_0$ is a hypothetical situation similar to $L(0,n,u_1)$ with the difference that the decision to use the subsequent AVF is delayed until $u_2+k$.
\begin{figure}[htbp]
\centering
\includegraphics[scale=0.60]{./files/fig11}
\caption{induction step - AVF creation successful}
\label{fig:fig11}
\end{figure}
\begin{itemize}
	\item $L(0,n,u_2) \le_{st} L_0$: The proof is the same as the proof we used for case 4 in the induction base case.
	\item $L_0 \le_{st} L(0,n,u_f1)$. For $x \le u_1+k$ the twfo random variables are identical. For $x  \ge u_1+k$, we use $L(u_1+k,n-1,u_2-u_1) \le_{st} L(u_1+k,n-1,0)$ by induction assumption to show the result.
\end{itemize}

\end{itemize}
\end{proof}


\section{Quality adjusted life expectation}
Using AVF for Hemo-dialysis not only brings better survival, but also has a slightly higher quality of life for the patient, in comparison with HD using CVC. Nevertheless, the process of AVF creation (AVF surgery) has some disutility associated with it, which can be attributed to the surgery and post-surgery inconveniences. We want to avoid this disutility if it cannot be compensated by the better survival and quality of life associated with HD on AVF. We define the patient's QALE as a weighted sum of their lifetime on each vascular access discounted by the AVF surgery disutility for each AVF operation they have done (whether successful or failure). The following parameters are used in defining the patient's QALE:
\begin{itemize}
\item $L_{A}$: RV denoting patient's total lifetime on AVF
\item $L_{C}$: RV denoting patient's total lifetime on CVC
\item $q_A$:   quality of life coefficient of HD using AVF
\item $q_C$:   quality of life coefficient of HD using CVC
\item $d$: AVF creation disutility
\item $Q$: RV denoting the patient's total QALE
\item $W$: RV denoting the patient's weighted lifetime
\item $N$: RV denoting number of surgeries done on the patient
\end{itemize}

\noindent Using these notations, we have the following equality for patient's total QALE under an arbitrary policy $\pi$:
\begin{align*}
Q^{\pi}=W^{\pi}-dN^{\pi}=q_A L_A^{\pi}+q_C L_C^{\pi}-dN^{\pi}
\end{align*}
Under the following assumption, we will show that the optimal referral policy to maximize patient's expected QALE is of threshold type as follows: for a given patient's age, for values of AVF creation disutility less than a critical value, i.e. $d \le d_{cr}$, it is optimal to refer patient for AVF creation at the time of decision; otherwise the optimal policy is to use CVC for the rest of patient's life, or in other words no more AVF creations.

Note that the critical value is dependant on the patient's age and the number of AVF chances left. More specifically, we will show that it is increasing in patient's age, and decreasing in number of AVF chances left.

\begin{assu} \label{ass:IFR}
The lifetime of patient on CVC, $C(0)$, has the increasing failure rate property. In mathematical words, we have that $\hr{g}{t}$ is increasing in $t$.
\end{assu}
Note that Assumption \ref{ass:IFR} and Assumption \ref{ass:converging} altogether imply the IFR property for the lifetime on AVF, $A(0)$. Also, based on Note \ref{note:origin}, we can show that $A(t)$, and $C(t)$ have IFR property for all values of $t$. The other implication is that $A(t_2) \le_{st} A(t_1)$ and $C(t_2) \le_{st} C(t_1)$ whenever $t_1 \le t_2$, which translates to the fact that a patient has a better survival on either of these accesses at younger ages.\\

In order to prove this claim, we set up a dynamic programming model using the following notations:
\begin{itemize}
\item $\pi_0$: policy of using CVC for the rest of the patient's life
\item $\pi(T)$: the threshold policy with threshold vector $T=(\tau_1, \ldots, \tau_n)$
\item $v^\pi(t,n)$: the value function (the expected residual QALE of patient) at time $t$ having $n$ remaining AVF chances following policy $\pi$ for AVF referral
\item $v(t,n)$: the optimal value function at time $t$ having $n$ remaining AVF chances
\end{itemize}

In order to prove the claim, we need to prove some preliminary results. We show that the difference of any feasible threshold policy $\pi(T)$ and the policy of no more referral ($\pi_0$) decreases as the patient ages. A feasible threshold policy $\pi(T)$ is a policy whose performance is no worse than that of $\pi_0$. We can easily show that a threshold policy is feasible iff $v^{\pi(T)}(t,1) \ge v^{\pi_0}(t,1)$. Thus, we want to prove that
\begin{align*}
\diff{v^{\pi(T)}(t,n)}{t} \le \diff{v^{\pi_0}(t,n)}{t}=\diff{v^{\pi_0}(t,0)}{t}: \forall n.
\end{align*}
given that $\pi(T)$ is feasible. We show that by means of induction. We first prove the induction base in the following lemma.
\begin{lem} \label{lem:thresh_dec_1}
Fix a threshold policy, and assume that the patient has one AVF chance left ($n=1$). Then the difference of value function of $\pi(T)$, and $\pi_0$ is decreasing in time.
\end{lem} 

\begin{proof}
If $\tau_1 <t$, then $\pi(T) =\pi_0$, and the result holds vacuously. Now, consider $t \le \tau_1$. Fix $M_1=m$, and $K_1=k$. We show that for all values of $m$, we have that 
\begin{align} \label{eq: dec_in_k1}
\diff{ \diff{v^{\pi(T)}(t,1|M_1, K_1=k)}{t}}{k} \le 0.
\end{align}
Showing this we have, 
\begin{align*}
\diff{v^{\pi(T)}(t,1|M_1)}{t} \le \diff{v^{\pi(T)}(t,1|M_1=m, K_1=0)}{t} = \diff{v^{\pi_0}(t,1)}{t}. 
\end{align*}
It remains to prove inequality \ref{eq: dec_in_k1}. But before that note that
\begin{align*} 
v^{\pi(T)}(t,1|M_1, K_1)=&-d+pv^{\pi(T)}(t,1|M_1, K_1,B_1=1)+(1-p)v^{\pi(T)}(t,1|M_1, K_1,B_1=0).
\end{align*}
Since $\diff{v^{\pi(T)}(t,1|M_1, K_1,B_1=0)}{k}=0$, we have that
\begin{align*} 
\diff{v^{\pi(T)}(t,1|M_1, K_1)}{k}=p\diff {v^{\pi(T)}(t,1|M_1, K_1,B_1=1)}{k}.
\end{align*}
Let $t':=t+m$. Note that we have
\begin{align*} 
v^{\pi(T)}(t,1|M_1, K_1,B_1=1)=-d+& q_C\int_{0}^{m} x\frac{g(x+t)}{\Gbar(t)}dx +\frac{\Gbar(t')}{\Gbar(t)}\bigg [q_Cm+\\
&q_A\int_{0}^{k} x \frac{h(x+t')}{\Hbar(t')} dx+\frac{\Hbar(t'+k)}{\Hbar(t')}\big[q_Ak+q_C\int_{t'+k}^{\infty} \frac{\Gbar(x)}{\Gbar(t'+k)} dx\big ] \bigg].
\end{align*}
Thus,
\begin{align*} 
\diff {v^{\pi(T)}(t,1|M_1, K_1,B_1=1)}{k}=&\bigg[\frac{\Gbar(t')}{\Gbar(t)}\bigg]\bigg[
q_Ak\frac{h(k+t')}{\Hbar(t')}+\frac{\Hbar(t'+k)}{\Hbar(t')} \bigg\{ q_A+q_C \big [-1+\frac{g(t'+k)}{\Gbar(t'+k)}\int_{t'+k}^{\infty} \frac{\Gbar(x)}{\Gbar(t'+k)} dx \bigg\}\\
&-\frac{h(t'+k)}{\Hbar(t')}  \bigg\{ q_Ak+q_C\int_{t'+k}^{\infty} \frac{\Gbar(x)}{\Gbar(t'+k)} dx \bigg \}
\bigg]\\
&=\bigg[\frac{\Gbar(t')}{\Gbar(t)}\bigg]\bigg[\frac{\Hbar(t'+k)}{\Hbar(t')}\bigg] \bigg[q_A-q_C+q_C\big[ \int_{t'+k}^{\infty} \frac{\Gbar(x)}{\Gbar(t'+k)} dx\big ] \big [\hr{g}{t'+k}-\hr{h}{t'+k}\big] \bigg]
\end{align*}

We can show that each of the terms in the brackets are positive and decreasing in $t$, and thus their product also has the same property. As a result we have that $\diff {v^{\pi(T)}(t,1|M_1, K_1)}{k}$ is decreasing in $t$. By switching the order of differentiation, we have the inequality \ref{eq: dec_in_k1}.
\end{proof}
Now, we prove the induction step by showing the following inequality:
\begin{align} \label{eq:relative}
\diff{v^{\pi(T)}(t,n)}{t} \le \diff{v^{\pi(T)}(t,n-1)}{t}
\end{align}
\begin{lem} \label{lem:thresh_dec_n}
Fix a threshold policy $\pi(T)$, and assume that the patient has $n$ AVF chance left. Then the inequality \ref{eq:relative} holds.
\end{lem}
\begin{proof}
~\\
Fix $M_i=m_i$, and $K_i=k_i$ for $i=2,\ldots,n$. If for some $j$ we have that $t+\sum_{i=j+1}^{n} m_i+k_i \ge \tau_j$; then we don't use any left AVF chances, and thus we have that $v^{\pi(T)}(t,n|M,K)=v^{\pi(T)}(t,n-l|M,K)$ for all $l \le j$. As a result the inequality \ref{eq:relative} holds in equality.\\
\indent Now assume that $t+\sum_{i=2}^{n} m_i+k_i \le \tau_1$, so referral happens at time $t+t'$, where $t':=\sum_{i=2}^{n} m_i+k_i$, provided that patient survives until that time; then the difference in the two cases is the (possible) use of one AVF chance. Let $S(t,t')$ represent the probability of survival of patient until time $t'$ in this scenario. Then we have the following:
\begin{align}
 v^{\pi(T)}(t,n|M,K)-v^{\pi(T)}(t,n-1|M,K) =\big[S(t',t) \big] \big[v^{\pi(T)}(t',1)-v^{\pi_0}(t',1)\big].
\end{align}
We prove that the LHD is decreasing in $t$ by showing that the terms in brackets are each non-negative and decreasing in $t$:
\begin{itemize} 
\item $S(t',t)$: This probability of survival is non-negative by definition. First note that we can compute $S(t',t)$ as follows:
\begin{align*}
S(t',t)=&\big[ P(C(t)>m_n)\big]\big[ P(A(t+m_1>k_n|C(t)>m_1)\big]\ldots \big[ P(A(t+t'-k_2)>k_2|C(t)>m_n ,\ldots)\big]\\
       =&\big[ P(C(t)>m_n) \big] \big[ P(A(t+m_1>k_n)\big] \ldots \big[ P(A(t+t'-k_2)>k_2)\big]
\end{align*}
each of the terms in the brackets are decreasing because of the IFR property of $A(.)$, and $C(.)$.
\item d
\end{itemize}
\end{proof}
~




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Is this true??
\begin{lem}\label{lem:cond}
Consider random variables $X,Y$ such that $X \le_{st} Y$. Then for all $a$, we have that 
\begin{align} \label{eq:conditional}
[X | X \ge a] \le_{st} [Y | Y \ge a].
\end{align}
\end{lem}
No, this is wrong \Frowny{}. According to \cite{shaked2007stochastic}, page 17, the equation \ref{eq:conditional} is equivalent to $X \le_{hr} Y$ which is not implied by $X \le_{st} Y$.


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